7 circuits, transmission lines, and waveguidesfab.cba.mit.edu/classes/862.16/notes/circuits.pdf ·...

7 Circuits, Transmission Lines, and Waveguides

Electric and magnetic fields contain energy, which can propagate. These are the ingre-

dients needed for communications; in this chapter we will look at how electromagnetic

energy can be guided. We will start with low-frequency circuits, then progress through

transmission lines to high-frequency waveguides.

7.1 C IRCUITS

The elements of an electrical circuit must satisfy Maxwell’s equations. In the low-

frequency limit this provides a fundamental explanation for the familiar circuit equations.

These simple relationships will hold as long as the frequencies are low enough for the size

of the circuit to be much smaller than the electromagnetic wavelength. Above this there

is a tricky regime in which the entire circuit acts like a distributed antenna, and then

when the wavelength becomes small compared to the size of the circuit things become

simpler again (this is the subject of Chapter 9 on optics).

7.1.1 Current and Voltage

The voltage or potential difference between two parts of a circuit is defined by the line

integral of the electric field

V = −∫

~E · d~l . (7.1)

As long as d ~B/dt = 0 then∇× ~E = 0, which implies that the electric field is the gradientof a potential and the value of its line integral is independent of the path; it can go through

wires or free space as needed and will always give the same answer. Conversely, if there

are time-varying magnetic fields then the potential difference does depend on path and

can no longer be defined as a function of position alone.

The electric field is defined to point from positive to negative charge so that the

potential increases along a path from negative to positive charge. A charge q such as anelectron in a wire feels a force ~F = q ~E, and so according to these definitions electronsflow from low to high potentials (Figure 7.1). The current ~I, in amperes, at a point ina wire is equal to the number of coulombs of charge passing that point per second. It is

defined to be in the same direction as the electric field and hence opposite to the direction

in which electrons travel. The current density ~J is equal to the current divided by itscross-sectional area ~J = ~I/A.

7.1 Circuits 81

V x

A

dx

s

L

E,J

F

Figure 7.1. A resistive element, showing the relations among the voltage, electric field,

current, and charge motion.

7.1.2 Kirchhoff’s Laws

There are two Kirchhoff Laws that can be used to analyze the current flow in a circuit:

• The sum of currents into and out of a circuit node must be zero.If multiple wires meet at a point, the sum of all their currents must be equal to

zero. This is just a statement of the conservation of charge.

• The sum of voltages around a circuit must vanish.This follows because the line integral of the electric field around a closed path

V = −∮

~E · d~l = −∫

∇× ~E · d ~A = ∂

∂t

∫

~B · d ~A (7.2)

will vanish if there is no time-varying magnetic flux linking the circuit.

7.1.3 Resistance

In an isotropic conductor the current and electric field are related by

~J = σ ~E , (7.3)

where σ is the material’s conductivity. For very large fields there may be nonlineardeviations from this linear relationship, and in a complex material the conductivity may

be a tensor that depends on direction. The voltage drop across the resistor in Figure

7.1 with length L, cross-sectional area A, conductivity σ, and carrying a current I istherefore

V = −∫ +

−

~E · d~x = −∫ +

−

~J

σ· d~x =

∫ +

−

I

σAdx =

IL

σA≡ IR . (7.4)

Remember that the integral goes from low to high potential, but that current flows from

high to low potentials, so − ~J · d~x = J dx = I dx/A. This is just Ohm’s Law, and itdefines the resistance

R =L

σA=

ρL

A(7.5)


in terms of the conductivity σ, which has units of siemens per meter (S/m), or the resis-tivity ρ, which has units of ohm-meters (Ω·m). For a two-dimensional film of thicknessT , the resistance of a region of length L and width W is

R = ρL

A

= ρL

TW

=ρ

T

L

W

≡ R⊔⊓L

W. (7.6)

This defines the sheet resistivity R⊔⊓ (“R square”). Since L/w is dimensionless, R⊔⊓ hasunits of resistance without any other length.

R2

V = IR1+IR

2

R1

V = IR2

V = 0

V

I

R2

V = I1R

1=I

2R

2

R1

V = 0

V

I2

I1

Figure 7.2. Series and parallel circuits.

Kirchhoff’s Laws can be used to simplify the circuits in Figure 7.2. For the series

circuit on the left,

V = IR1 + IR2 (7.7)

or

I =V

R1 + R2

, (7.8)

therefore the total resistance is

Rtotal = R1 + R2 . (7.9)

Series resistances simply add. For the parallel circuit on the right, the voltage drop across

both legs must be equal since potential is independent of path,

V = I1R1 = I2R2 , (7.10)

and the current in both legs must add up to the total current

I1 + I2 =V

Rtotal

. (7.11)

7.1 Circuits 83

ThereforeV

R1

+V

R2

=V

Rtotal

(7.12)

or1

Rtotal

=1

R1

+1

R2

Rtotal =R1R2

R1 + R2

. (7.13)

Parallel resistances add inversely. More complex networks of resistances can always be

simplified to a single effective resistance by repeated application of these rules.

7.1.4 Power

Now let’s now consider a slab of charge of cross-sectional areaA and thickness dxmovingthrough the resistor in Figure 7.1. If the charge density is ρq, the total charge in this slab

is Q = ρqdxA and it feels a net force ~F = Q~E. Because a current is flowing, charge ismoving relative to this force and so work is being done. The work associated with the

slab moving from one end of the resistor to the other is equal to the integral of the force

times the displacement:

dW =

∫ +

−

~F · d~x = −Q

∫ +

−

~E · d~x = −QV = −ρqdxAV (7.14)

for a negative charge. This decrease in energy is dissipated in the resistor; the power is

equal to the rate at which work is being done

P = −dW

dt= ρq

dx

dtAV = JAV = IV = I2R . (7.15)

The power dissipated in a resistor is equal to the current flowing through it times the

voltage drop across it, which by Ohm’s Law is also equal to the square of the current

times the resistance. This appears as heat in the resistor.

7.1.5 Capacitance

There will be an electric field between an electrode that has a charge of +Q on it and

one that has a charge of −Q, and hence a potential difference between the electrodes.Capacitance is defined to be the ratio of the charge to the potential difference:

C =Q

V. (7.16)

The MKS unit is the farad, F. Capacitances range from picofarads in circuit components

up to many farads in supercapacitors based on electrochemical effects [Conway, 1991].

The current across a capacitor is given by

CdV

dt=

dQ

dt= I . (7.17)

A capacitor is a device that stores energy in an electric field by storing charge on its plates;

in Problem 6.2 we saw that this stored energy is equal to CV 2/2. The current flowing


across a capacitor is a displacement current: from the point of view of the overall circuit

it is a real current, but it arises from the time-varying electric field associated with the

capacitor plates storing or releasing charge rather than from real charge passing through

the capacitor.

If the applied voltage is V = eiωt, then the current is

I = CdV

dt= iωCeiωt . (7.18)

The impedance (complex resistance) is defined to be the ratio of the voltage and current

at a fixed frequency,

Z =V

I=

eiωt

iωCeiωt=

1

iωC. (7.19)

The current leads the voltage by a phase shift of i = 90. When ω = 0 the impedanceis infinite (no current flows at DC), and when ω =∞ the impedance is 0 (the capacitor

acts like a wire).

7.1.6 Inductance

An inductor stores energy in a magnetic field arising from current flowing through a coil.

The inductance is defined to be the ratio of the magnetic flux

Φ =

∫

~B · d ~A (7.20)

linking a circuit to the current that produces it:

L =Φ

I. (7.21)

The MKS unit of inductance is the henry, H.

In Figure 7.3 the electric field vanishes along the dotted line for an ideal solenoid,

therefore the line intergral of the electric field along the dotted line and around the

solenoid is equal to the voltage drop across the solenoid. And the magnetic field vanishes

outside the solenoid if it is approximated to be a section of an infinite solenoud, therefore

the integral of the magnetic field across the surface bounded by the path is equal to the

flux linking the solenoid times the number of turns of the coil. This lets us relate the

flux to the potential. If this ideal solenoid is taken to have just a single turn then

−V =

∮

~E · d~l = − ∂

∂t

∫

S

~B · d ~A = −∂Φ

∂t= − ∂

∂t(LI) = −L

dI

dt(7.22)

and so

V = LdI

dt. (7.23)

Extra turns add in series: if an inductor has N turns, then the inductance is N times that

due to one turn, assuming that the flux linking all the turns is the same. Since the field

of a solenoid of radius r and length l with n turns/meter is H = nI, the inductance is

L =Φ

I=

N

I

∫

~B · d ~A = nl

IµnIπr2 = µn2lπr2 . (7.24)

7.2 Wires and Transmission Lines 85

I

Figure 7.3. A solenoid; the dotted line closes the integration path.

Problem 6.3 showed that the energy stored in a solenoid is LI2/2.

If the current flowing through an inductor is I = eiωt then the voltage drop across it

is V = Liωeiωt, and so the impedance is

Z =Liωeiωt

eiωt= iωL . (7.25)

The current lags the voltage by 90 (−i).

7.2 WIRES AND TRANSMISSION LINES

We have been considering conduction in the low-frequency limit; in this section we will

use Maxwell’s equations to look at how AC fields penetrate conductors and are guided

by them at higher frequencies.

7.2.1 Skin Depth

Assume that the conductor is described by ~J = σ ~E, ~D = ǫ ~E, ~B = µ ~H. If the electricfield is periodic as ~E(~x, t) = ~E(~x)eiωt then the curl of the magnetic field is

∇× ~H = ~J +∂ ~D

∂t

∇× ~B = µσ ~E + µǫ∂ ~E

∂t

∇× ~B(~x) = (µσ + iωµǫ)~E(~x) . (7.26)

Since the divergence of a curl vanishes,

∇ · ∇ × ~B = 0 = (µσ + iωµǫ)∇ · ~E

⇒ ∇ · ~E =ρ

ǫ= 0 . (7.27)

The linear response coefficients require that there be no free charge.

Now taking the curl of the curl of ~E,

∇× ~E = −∂ ~B

∂t


∇×∇× ~E = − ∂

∂t∇× ~B

∇ (∇ · ~E)︸︷︷︸

0

−∇2 ~E = −µσ∂ ~E

∂t− µǫ

∂2 ~E

∂t2. (7.28)

The first term on the right hand side is due to real conduction, and the second term is

due to the displacement current. Since σ is very large in a good conductor, up to veryhigh (optical) frequencies the displacement current term can be dropped:

∇2 ~E = µσ∂ ~E

∂t. (7.29)

This is now a diffusion equation instead of a wave equation. For a periodic electric field,

the spatial part satisfies

∇2 ~E(~x) = iωµσ ~E(~x) ≡ k2 ~E(~x) . (7.30)

Since√i =

1 + i√2

(7.31)

(try squaring it),

k =√

iωµσ

= (1 + i)

√ωµσ

2

≡ 1 + i

δ. (7.32)

This defines the skin depth

δ =

√

2

ωµσ=

1√πνµσ

(7.33)

in terms of the frequency ν, the permeability µ, and the conductivity σ.

Consider the solution to equation (7.30) for a plane wave incident on the surface of a

conductor, so that by symmetry we need consider only the distance z into the conductor

d2E

dz2= k2E . (7.34)

E is the magnitude of the electric field, which for a plane wave is transverse to the

direction of z. This is solved by

E(x) = E0e−kz = E0e

−z/δe−iz/δ , (7.35)

where E0 is the amplitude at the surface, and we’ve ignored the unphysical possiblesolution ekz . The total current per unit width that is produced by this field is found byintegrating the current density over the depth

I =

∫ ∞

0

J dz =

∫ ∞

0

σE dz =

∫ ∞

0

σE0e−kz dz =

σE0k

. (7.36)


Therefore

E0 =kI

σ=1 + i

σδI =

(1

σδ+ i

1

σδ

)

I

≡ (Rs + ωLs)I . (7.37)

The total current is proportional to the applied field at the surface; the real part of this

defines an effective surface resistance Rs, and the imaginary part defines the surface

inductance Ls. Associated with this current there is dissipation; in a small volume ofcross-sectional area A and length along the surface L the dissipation per volume is

I2volumeR

AL=

1

ALJ2A2

L

σA=

J2

σ. (7.38)

If the current is periodic, taking a time average introduces another factor of 〈sin2〉 = 1/2:⟨I2R

AL

⟩

=|J |22σ

. (7.39)

Integrating this from z = 0 to∞ gives the energy dissipated by the field in the material

per surface area∫ ∞

0

|J |22σ

dz =

∫ ∞

0

σ2E20

2σe2z/δ dz =

σE20δ

4. (7.40)

The amplitude of the field and current are falling off exponentially with a length scale

equal to the skin depth. For example, pure copper at room temperature has a conductivity

of 5.8 × 107 S/m and so δ ∼ 7 cm at 1 Hz, 2 mm at 1 kHz, 70 µm at 1 MHz, and 2

µm at 1 GHz. Since the skin depth is so small at even fairly low frequencies, very little

thickness is needed to screen a field. This is why it is a good approximation to assume

that fields vanish at the surface of a conductor, which we have already found to be the

boundary condition for a perfect conductor. The part of the field that does leak into

the conductor causes a current to flow, and this current leads to resistive dissipation,

therefore in making this approximation we are leaving out the mechanism that damps

fields around conductors. This is very important in resonant electromagnetic cavities that

are designed to have a high Q (low damping rate).

Because of the skin depth, a bundle of fine wires has a smaller AC resistance than a

single fat wire because there is more surface area for the current to penetrate into and

the overall resistance will be inversely proportional to the effective cross-secional area of

the bundle. This is why wires carrying high frequency signals are stranded rather than

solid.

7.2.2 Transmission Lines

While electromagnetic fields cannot penetrate far into good conductors, they can be

guided long distances by them. Distributed objects can have energy stored in electric

fields through capacitance, and in magnetic fields by inductance; the interplay between

these can give rise to an energy flow. As a first example such a transmission line, consider

the coaxial cable in Figure 7.4). Other important transmission line geometries include

parallel wires or strips (Problem 7.4), and a strip above a ground plane (called a stripline).

Because a transmission line is operated in a closed circuit there is no net charge transfer


I

-I

z

ri

ro

e

z

Figure 7.4. A coaxial cable field with a dielectric ǫ.

between either end. We will assume here that any current I in the inner conductor mustbe matched by a return current −I in the outer conductor; the next section will studyhigher-frequency modes for which this is no longer true. There is an electric field between

the inner and outer conductors, giving rise to a distributed capacitance between them.

Current flowing in the inner conductor also produces a magnetic field around it, and

hence a distributed inductance along it. As long as the frequency is not so large that

the wavelength is comparable to the cross-sectional size, the coaxial cable therefore acts

like an extended series inductor and parallel capacitor (Figure 7.5). This circuit model is

applicable to arbitrary transmission lines; its solution will reappear in the next section as

the fundamental mode for Maxwell’s equation in a cylindrical geometry.

I

C dz

L dz

V V+DV

dz

L dzI I+DI

I- I I- - D

I-

Figure 7.5. Effective circuit model for a transmission line, and a differential element.

From Stokes’ Law, the magnetic field between the conductors is

H =I

2πr, (7.41)

and the field vanishes outside of the outer conductor because the net current is then zero.

Integrating this field over the surface of length z shown in Figure 7.4 to find the fluxbetween the conductors,

Φ =

∫

~B · d ~A = z

∫ ro

ri

µ0I

2πrdr = z

µ0I

2πln

rori

. (7.42)

Since the dielectric is non-magnetic we can take µr ∼ 1. Therefore, the inductance per


length is

L = Φ

zI=

µ02πln

rori

(H

m

)

. (7.43)

Similarly, from Gauss’ Law the electric field between the conductors is

E =Q

2πǫr, (7.44)

whereQ is the charge per unit length and the field vanishes outside of the outer conductor.Integrating to find the potential difference,

V = −∫ ro

ri

~E · d~l = Q

2πǫln

rori

, (7.45)

which gives the capacitance per unit length

C = Q

V=

2πǫ

ln(ro/ri)

(F

m

)

. (7.46)

7.2.3 Wave Solutions

Now consider the little differential length of the transmission line dz shown in Figure7.5, with parallel capacitance C dz and series inductance L dz. If there is an increase inthe current flowing across it

∆I =∂I

∂zdz (7.47)

there must be a corresponding decrease in the charge stored in the capacitance

∆I = −Cdz ∂V∂t

. (7.48)

Therefore

∂I

∂zdz = −Cdz ∂V

∂t(7.49)

or

∂I

∂z= −C ∂V

∂t. (7.50)

Similarly, if there is an increase in the potential across the element

∆V =∂V

∂zdz (7.51)

there must be a decrease in the current flowing through the inductance

∆V = −Ldz ∂I∂t

. (7.52)

Current flows from high to low potential, so an increasing potential drop across the

inductor has the opposite sign from the decreasing current. Equating these expressions,

∂V

∂z= −L∂I

∂t. (7.53)


Now take a time derivative of equation (7.50)

∂2I

∂t∂z= −C ∂

2V

∂t2(7.54)

and a z derivative of equation (7.53)

∂2V

∂z2= −L ∂2I

∂z∂t(7.55)

and interchange the order of differentiation to equate the mixed terms (which is permitted

for well-behaved functions):

∂2V

∂z2= LC ∂

2V

∂t2≡ 1

v2∂2V

∂t2, (7.56)

where

v ≡ 1√LC

. (7.57)

This is a wave equation for the voltage in the transmission line. It is solved by an arbitrary

distribution traveling with a velocity ±v

V (z, t) = f (z − vt) + g(z + vt) (7.58)

= V+ + V− .

If we follow a fixed point in the distribution f (0), z−vt = 0⇒ z = vt. The V+ solutiontravels to the right, and V− to the left. For a sinusoidal wave V = ei(kz−ωt), k = ω/v.Taking derivatives in the opposite order gives a similar equation for the current:

∂2I

∂z2=1

v2∂2I

∂t2. (7.59)

To relate the voltage to the current, substitute equation (7.58) into equation (7.50)

∂I

∂z= −C[−vf ′(z − vt) + vg′(z + vt)] (7.60)

and integrate over z

I = Cv[f (z − vt) − g(z + vt)]

≡ 1

Z[f (z − vt)− g(z + vt)]

=1

Z[V+ − V−]

= I+ + I− , (7.61)

where

Z =1

Cv =√

LC (Ω). (7.62)

The current is proportional to the voltage, with the sign difference in the two terms

coming from the difference between the solutions traveling in the right and left directions.

The constant of proportionality is the characteristic impedance of the transmission line

Z. The velocity and impedance of a transmission line are simply related to the capacitanceand inductance per unit length. In a real cable, different frequencies are damped at


different rates, changing the pulse shape as it travels, and if there are nonlinearities

then different frequencies can travel at different rates causing dispersion: a sharp pulse

will spread out. The dispersion sets a limit on how close pulses can be and still remain

separated after traveling a long distance.

7.2.4 Reflections and Terminations

Consider a transmission line with a characteristic impedance Z0 terminated by a loadimpedance ZL. The load might be a resistor, or it could be another transmission line.For a resistor the impedance is associated with energy dissipated by ohmic heating, and

for a transmission line the impedance is associated with energy that is transported away,

but in both cases the voltage drop across the element is equal to the current applied to it

times its impedance.

The incoming transmission line can support signals traveling in both directions (equa-

tion 7.58), therefore the voltage at the discontinuity is the sum of these:

VL(t) = V+(t) + V−(t) . (7.63)

Similarly, the current across the load is

IL(t) = I+(t) + I−(t) . (7.64)

The current across the termination must equal the current in the transmission line im-

mediately before the termination:

VLZL

=V+Z0

− V−

Z0. (7.65)

Eliminating variables between this and equation (7.63) gives the ratio of the incoming

and reflected voltages, called the reflection coefficient

R =V−

V+=

ZL − Z0ZL + Z0

, (7.66)

and the ratio of the incoming and the transmitted signals is equal to the transmission

coefficient

T =VLV+=

2ZLZL + Z0

. (7.67)

Because of the load, V+ and V− can no longer be arbitrarily chosen but must satisfy

the boundary conditions. These reflection and transmission coefficients have a number

of interesting properties. If the load impedance is 0 (a short), R = −1 and so there is areflected pulse of the same shape but opposite sign. If the load resistance is infinite (it

is open), R = 1 and the reflected pulse has the same sign. These reflections are used

in a Time Domain Reflectometer (TDR) to locate cable faults by measuring the time

for a return pulse to arrive. Finally, if ZL = Z0 then R = 0: there is no reflection at

all! This is why cables carrying high-frequency signals are terminated with resistors that

match the cable’s characteristic impedance. Such terminations are particularly important

to eliminate clutter from reflected pulses in computer networks and buses.


If V+(z) = V0eikz going into the load,

V (z) = V+(z) + V−(z)

= V0(eikz + Re−ikz

). (7.68)

As a function of z, the positive- and negative-going waves will periodically add to andsubtract from each other. Taking the ratio of the maximum to the minimum value for

this sum defines the Voltage Standing-Wave Ratio (VSWR)

VSWR ≡ VmaxVmin

=1 + |R|1− |R| (7.69)

or

|R| = VSWR − 1VSWR + 1

. (7.70)

The VSWR is one of the most important measurements in an RF system, used to ensure

that impedances are matched so that all of the power goes in the intended direction.

Z0

ZL

z–d

VL,I

L

V+ ,I+

V–

,I–

Z

Figure 7.6. A transmission line with impedance Z terminated by a load ZL.

Now consider the impedance of a transmission line as viewed by a periodic source a

distance d from the termination, shown in Figure 7.6:

Z(−d) =V (−d)

I(−d)

=V+e

−ikd + V−eikd

Z−10

(V+eikd − V−e−ikd

)

=V+

(e−ikd +Reikd

)

V+Z−10

(eikd − Re−ikd

)

= Z0

(e−ikd +Reikd

)

(eikd −Re−ikd

) . (7.71)

Normalizing this by the characteristic impedance of the transmission line,

Z(−d)

Z0=

e−ikd +Reikd

e−ikd −Reikd


=1 + Rei2kd

1−Rei2kd

r + ic ≡ 1 + (x + iy)

1− (x + iy)

=1− (x2 + y2)

(1− x)2 + y2)+ i

2y

(1− x)2 + y2, (7.72)

relates the real and complex parts of the input impedance

r + ic =Z

Z0(7.73)

to those of the round-trip reflection coefficient

x + iy = Rei2kd . (7.74)

The real equation can be rewritten suggestively as

r =1− (x2 + y2)

(1− x)2 + y2

r(1 − x)2 + x2

1 + r+ y2 =

1

1 + r

x2 − 2x r

1 + r+

r

1 + r+ y2 =

1

1 + r

x2 − 2x r

1 + r+( r

1 + r

)2

+ y2 =1

1 + r+( r

1 + r

)2

− r

1 + r

(

x− r

1 + r

)2

+ y2 =1

(1 + r)2. (7.75)

In the complex (x, y) plane, the reflection coefficient lies on a circle of radius 1/(1 + r)with a center at (r/(1 + r), 0) set by the real part of the input impedance r. Similary, thecomplex equation can be rewritten as

c =2y

(1− x)2 + y2

(1− x)2 + y2 =2y

c

(1− x)2 + y2 − 2y 1c+1

c2=1

c2

(1− x)2 +

(

y − 1

c

)2

=1

c2. (7.76)

This restricts the reflection coefficient to a circle of radius 1/c located at (1,±1/c) givenby the complex part of the input impedance c. The intersection of these two circlesrelates the input impedance to the reflection coefficient, conveniently found graphically

on a Smith chart (Figure 7.7).


r = 0

r = 0.5r = 1

r = 2

c = 1

c = 2

c = 0.5

c = -0.5

x

y

|R| = 1

c = 0

90°

-

135°

-1c = -

2c = -

90°

45°

135°

0°180

°

45°

-

Figure 7.7. The Smith chart.

7.3 WAVEGUIDES

As the wavelength of a signal in a transmission line becomes comparable to the transverse

size of the line, more complicated excitations become possible and the circuit model used

in the last section no longer applies. A complete solution of Maxwell’s equations is

then required. Some of these new modes will prove to be desirable, and some will not.

Waveguides, not surprisingly, guide electromagnetic waves. Depending on the geometry

they may or may not be able to transmit a steady current because it is possible to guide

waves without a DC return path. Waveguides usually have some symmetry about their

long axis; a rectangular pipe is a common type.

7.3.1 Governing Equations

Start with the wave form of Maxwell’s equations without any sources:

∇2 ~E = µǫ∂2 ~E

∂t2∇2 ~H = µǫ

∂2 ~H

∂t2. (7.77)

We are looking for waves that travel along the axis of the waveguide periodically as

eiωt−γz. The real part of γ is the decay rate of the wave and the complex part is the wavevector 2π/λ. Cancelling out the time dependence,

∇2 ~E = −ω2µǫ~E ≡ −k2 ~E ∇2 ~H = −k2 ~H . (7.78)

7.3 Waveguides 95

The Laplacian can be separated into components that are transverse to the waveguide

axis and that are axial, taken here to be in the ~z direction:

∇2 ~E = ∇2T~E +

∂2 ~E

∂z2

= ∇2T~E + γ2 ~E . (7.79)

This turns equations (7.78) into Helmholtz’ equations for the transverse dependence of

the field

∇2T~E = −(γ2 + k2)~E ≡ −k2c

~E ∇2T~H = −k2c

~H , (7.80)

defining the characteristic wave vector kc. Along with this, the curl equation for a periodicsignal

∇× ~E = −iωµ ~H (7.81)

has transverse components

∂Ez

∂y− ∂Ey

∂z︸︷︷︸

−γEy

= −iωµHx∂Hz

∂y− ∂Hy

∂z︸︷︷︸

−γHy

= iωµEz

∂Ex

∂z︸︷︷︸

−γEx

−∂Ez

∂x= −iωµHy

∂Hx

∂z︸︷︷︸

−γHx

−∂Hz

∂x= iωµEy (7.82)

which can be rearranged as

Ex = − 1

k2c

(

γ∂Ez

∂x+ iωµ

∂Hz

∂y

)

Hx =1

k2c

(

iωǫ∂Ez

∂x− γ

∂Hz

∂y

)

Ey =1

k2c

(

−γ∂Ez

∂x+ iωµ

∂Hz

∂y

)

Hy = − 1

k2c

(

iωǫ∂Ez

∂x+ γ

∂Hz

∂y

)

. (7.83)

If the axial components Ez, Hz are found from equations (7.80), they completely deter-

mine the transverse components through equations (7.83).

This set of equations admits three kinds of solutions: Transverse Electric (TE) with

Ez = 0, Transverse Magnetic (TM) with Hz = 0, and Transverse Electromagnetic

(TEM) with Ez = Hz = 0. For the TEM case, because the numerator in equations

(7.83) vanishes, the only way the transverse components can be non-zero is for the

denominator k2c = γ2+k2 to also vanish. This means that γ = ±ik = ±iω√µǫ = ±iω/c,

therefore TEM waves travel at the speed of light in the medium. kc = 0 also reducesHelmholtz’ equations to Laplace’s equation, giving the static field solutions we used

when studying transmission lines. Because in a hollow conductor the boundary is an

equipotential, Laplace’s equation implies that the field must vanish everywhere in the

interior, therefore a TEM wave cannot be supported. Adding another conductor, such

as the center lead in a coaxial cable, makes a TEM solution possible.


7.3.2 Rectangular Waveguides

Now consider a rectangular waveguide with width w in the x direction and height h inthe y direction. The transverse equation for a TM wave is

∇2TEz =

∂2Ez

∂x2+∂2Ez

∂y2= −k2cEz . (7.84)

Solving this subject to the boundary condition that the field must vanish at the conducting

surfaces at x = 0, w and y = 0, h gives

Ez = A sin(kxx) sin(kyy) , (7.85)

where

k2c = k2x + k2y

kxw = mπ

kyh = nπ (7.86)

index the possible modes as a function of integers m and n. If we define a characteristicfrequency ωc associated with each mode by

ωc(m,n) =kc(m,n)√

µǫ=

1√µǫ

[(mπ

w

)2

+(nπ

h

)2]1/2

, (7.87)

then we can find the propagation constant

γ2 = k2c − k2

= k2c

(

1− k2

k2c

)

= k2c

(

1− ω2µǫ

ω2cµǫ

)

= k2c

(

1− ω2

ω2c

)

. (7.88)

Therefore

γ = kc(m,n)

[

1− ω2

ωc(m,n)2

]1/2

ω < ωc(m,n)

γ = ikc(m,n)

[(ω

ωc(m,n)

)2

− 1]1/2

ω > ωc(m,n) . (7.89)

When ω is less than the cutoff frequency ωc for a mode, or equivalently when thewavelength λ is greater than the cut-off wavelength λc, γ is pure real and so the modedecays exponentially. When ω is greater than the cut-off frequency for a mode, γ is pureimaginary and the mode propagates. These modes are labeled TMmn. Repeating this

analysis for the TE wave by starting with the transverse equation for Hz shows that the

TE and TM waves are degenerate with the same cutoff frequencies. At low frequencies

nothing propagates; as the frequency is raised more and more modes can be excited, with

the distribution of energy among them depending on how the waveguide is driven.

7.3 Waveguides 97

7.3.3 Circular Waveguides

For a waveguide with cylindrical symmetry, the transverse Laplacian for a TM mode is

∇2TEz =

1

r

∂

∂r

(

r∂Ez

∂r

)

+1

r2∂2Ez

∂θ2= −k2cEz , (7.90)

which is solved by Bessel functions of the first (Jn) and second (Nn) kind [Gershenfeld,

1999a]:

Ez(r, θ) = [AJn(kcr) +BNn(kcr)][C cos(nθ) +D sin(nθ)] . (7.91)

The modes TMnl are indexed by the order of the Bessel function n, and the root lof the Bessel function needed to make the field vanish at the boundaries. Although

these frequencies can no longer be solved for analytically, for a coaxial cable a rough

approximation for the TM modes is to ask that the wavelength be a multiple of radial

spacing

λc ≈2

n(ro − ri) n = 1, 2, 3, . . . , (7.92)

and for a TE mode that there be an integer number of azimuthal cycles

λc ≈2π

n

a + b

2(7.93)

[Ramo et al., 1994]. In the section on transmission lines we studied the fundamental

TEM mode. Because these higher-order modes have different velocities, if they are

excited they will spread out the signal and hence limit the usefulness of the cable. This

is why waveguides are usually designed to be operated with a single mode.

7.3.4 Dielectric Waveguides and Fiber Optics

Fortunately for telecommunications, waves can be guided by dielectric rather than con-

ducting waveguides. The surface resistance that we saw in Section 7.2.1 represents a

significant drag on a wave traveling in a waveguide, limiting the distance over which it

is useful. Also, the requirement that the transverse dimension of a guide be compara-

ble to the wavelength that is carried is easily met at microwave frequencies from ∼1 to100 GHz (∼10 cm to 1 mm), but it becomes impractical at higher frequencies to workwith macroscopic objects with microscopic dimensions. Both of these problems can be

addressed by carrying light in a glass fiber instead of RF in a metal box.

To see how a wave can be guided by dielectrics, consider the slab geometry shown in

Figure 7.8. We’ll look for a mode confined in the y direction with a periodic z dependenceof e−γz ≡ e−iβz. Starting with the TE mode, the transverse equation for Hz becomes

d2H

dy2= −(γ2 + k2)Hz = (β

2 − k2)Hz (7.94)

because there is no variation in the x direction. Depending on the relative magnitudes ofβ and k this can have oscillatory or exponential solutions. For the solution to be confined,and reflect the symmetry of the structure, we require the wave to be exponentially damped


x

y z

h

h-

e0

e1

e0

Figure 7.8. A dielectric slab waveguide.

outside of the slab, and periodic across it:

d2Hz

dy2=

(β2 − k20

)Hz

(|y| > h

)

−(k21 − β2

)Hz

(|y| < h

) . (7.95)

The symmetric solution to this is

Hz =

Ae−(β2−k20)

1/2(|y|−h) ≡ Ae−a(|y|−h)(|y| > h

)

B cos((k21 − β2)1/2y

)≡ B cos(by)

(|y| < h

) . (7.96)

Now the boundary conditions require continuity of the field at the interfaces, hence

A = B cos(bh) . (7.97)

The transverse components are found from equations (7.83), which for Ex is

Ex = −iωµ

k2c

∂Hz

∂y

= −iωµ

k2 − β2∂Hz

∂y

=

−iωµa Ae−a(y−h) (y > h)

iωµbB sin(by) (|y| < h)

iωµa Ae−a(−y−h) (y < −h)

. (7.98)

Equating these again at the boundaries,

A

a=

B

bsin(bh) . (7.99)

Now divide equation (7.99) by (7.97) to find

1

a=1

btan(bh) . (7.100)

This is a transcendental equation relating a and b, with multiple branches because of the

7.4 Selected References 99

periodicity of tan(bh). A second relationship comes from the definitions of a and b

a2 = β2 − k20

b2 = k21 − β2

⇒ a2 + b2 = k21 − k20 . (7.101)

a and b are restricted to a circle, with a radius given by the difference of the squaresof k2 = ω2µǫ in the media. For a and b to be real, the central slab must have thehigher dielectric constant. The intersections of these circles with the branches of equation

(7.100), found graphically or numerically, give the modes of the waveguide.

The analysis is similar for rectangular slabs that confine modes in both directions and

for circular dielectric waveguides, although the imposition of these boundary conditions

becomes a more difficult calculation [Yariv, 1991]. The result for the circular geometry

is that there are two modes with axial H and E components, one called the HE with Hdominant, and an EH mode with E dominant.

Dielectric waveguides for confining light are produced by depositing core doping

material on the inside of a cladding glass tube and then drawing it down to a thin optical

fiber. The first ones were multi-mode fibers that had core diameters many times the

optical wavelength, resulting in very dispersive communications. In the next chapter

we’ll see that this can be understood as many different path lengths reflecting at the

core–cladding interface. Because they’re easier to make and connect to, these are still

are used for short links and for many optical sensors that measure light coupling into

or out of a fiber to determine local material properties [Merzbacher et al., 1996], but

long-haul communications uses single-mode fibers. The minimum absorption in optical

glasses occurs at infrared wavelengths; by using very pure materials this has been reduced

below 0.2 dB/km at 1.55 µm [Miya et al., 1979; Takahashi, 1993]. This corresponds to

a loss of 10−3 over 150 km, making long links possible without active repeaters.

So far we’ve been considering step-index fibers that have a constant dielectric constant

in the core. By varying the core doping as a function of thickness it’s possible to make

graded-index fibers that use the radial profile to shape the modes. And an asymmetrical

blank when drawn down produces a polarization-preserving fiber that retains the po-

larization of the light [Galtarossa et al., 1994]. We’ve also assumed that the medium is

linear, but the intense fields in the small fiber cores can excite nonlinear effects. We’ll

see more of this in the Chapter 9, but one of the most important applications is to the

creation of solitons [Zabusky, 1981]. These are pulses that balance the material’s intrinsic

frequency-dependent dispersion with a nonlinear response that narrows the pulse, result-

ing in a stable shape that can propagate for long distances without changing. These can be

sent across ocean-scale distances at Gbits/second without errors [Nakazawa et al., 1993;

Mollenauer et al., 1996]. Using all of these tricks, fiber links have been demonstrated at

speeds above 1 Tbit/second, approaching the limit of 1 bit/second per hertz of optical

bandwidth [Ono & Yano, 1998; Cowper, 1998].

7.4 SELECTED REFERENCES

[Ramo et al., 1994] Ramo, Simon, Whinnery, John R., & Duzer, Theodore Van. (1994).Fields and Waves in Communication Electronics. 3rd edn. New York: Wiley.


A nice introduction to applied electromagnetics.

[Hagen, 1996] Hagen, Jon B. (1996). Radio-Frequency Electronics: Circuits andApplications. New York: Cambridge University Press.

[Danzer, 1999] Danzer, Paul (ed). (1999). The ARRL Handbook for Radio Amateurs.76th edn. Newington, CT: American Radio Relay League.

Practical details for all aspects of RF design.

7.5 PROBLEMS

(7.1) Cables designed to carry a low-frequency signal with minimum pickup of interfer-

ence often consist of a twisted pair of conductors surrounded by a grounded shield.

Why the twist? Why the shield?

(7.2) Salt water has a conductivity ∼ 4 S/m. What is the skin depth at 104 Hz?(7.3) Integrate Poynting’s vector ~P = ~E × ~H to find the power flowing across a cross-

sectional slice of a coaxial cable, and relate the answer to the current and voltage

in the cable.

(7.4) Find the characteristic impedance and signal velocity for a transmission line con-

sisting of two parallel strips with a width w and a separation h (Figure 7.4). Youcan ignore fringing fields by assuming that they are sections of conductors infinitely

wide.

w

h

I

I-

Figure 7.9. Transmission line for Problem 7.4.

(7.5) The most common coaxial cable, RG58/U, has a dielectric with a relative permit-

tivity of 2.26, an inner radius of 0.406 mm, and an outer radius of 1.48 mm.

(a) What is the characteristic impedance?

(b) What is the transmission velocity?

(c) If a computer has a clock speed of 1 ns, how long can a length of RG58/U be

and still deliver a pulse within one clock cycle?

(d) It is often desirable to use thinner coaxial cable to minimize size or weight but

still match the impedance of RG58/U (to minimize reflections). If such a cable

has an outer diameter of 30 mils (a mil is a thousandth of an inch), what is the

inner diameter?

(e) For RG58/U, at what frequency does the wavelength become comparable to

the diameter?

7.5 Problems 101

(7.6) Consider a 10 Mbit/s ethernet signal traveling in a RG58/U cable.

(a) What is the physical length of a bit?

(b) Now consider what would happen if a “T” connector was used to connect one

ethernet coaxial cable to two other ones. Estimate the reflection coefficient for

a signal arriving at the T.

7 circuits, transmission lines, and waveguidesfab.cba.mit.edu/classes/862.16/notes/circuits.pdf ·...

Documents